1. The azimuths of the back andforward tangents of a simple curverespectively are 257° 25' and 294005. The long chord of the simplecurve is 123.147 meters andstationing of V is at station 50 + 134.a. Determine the other elements ofthe simple curve. Use arc definition.b. Determine how many intermediatestations are in between station PCand Station PT.c. Compute the deflection anglesneeded for the layout of this simplecurve

Given:- Azimuths of the back tangents of a simple curve =257° 25' andAzimuths of forward tangents=…

1. The azimuths of the back and forward tangents of a simple curve respectively are 257° 25' and 2940 05. The long chord of the simple curve is 123.147 meters and stationing of V is at station 50 + 134.

1. The azimuths of the back and forward tangents of a simple curve respectively are 257° 25' and 2940 05. The long chord of the simple curve is 123.147 meters and stationing of V is at station 50 + 134.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

1. The azimuths of the back and
forward tangents of a simple curve
respectively are 257° 25' and 2940
05. The long chord of the simple
curve is 123.147 meters and
stationing of V is at station 50 + 134.
a. Determine the other elements of
the simple curve. Use arc definition.
b. Determine how many intermediate
stations are in between station PC
and Station PT.
c. Compute the deflection angles
needed for the layout of this simple
curve

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